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If a copper wire is stretched so that its length increases by `20%` then what is the percentage increase in its resistance (assuming its volume remaing constant) ?A. `10%`B. `21%`C. `44%`D. `120%` |
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Answer» Correct Answer - C `(R)/(R_(o))=((l)/(l_(o)))^(2)=((1.2l)/(l))^(2)` `R_(f)=1.44R_(o)` `therefore%"Changing in resistance"=(1.44R_(o))/(R_(o))xx100=0.44xx100=44%.` |
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